Question

Срочно !!!!
Пожалуйста решите уравнения.

Answer 1

а) $frac{x^{2}+10x}{10} - 20 = frac{2x+5}{2}$

$frac{x^{2}+10x}{5}$ - 40 = 2x + 5

$x^{2}$ + 10x - 200 = 10x + 25

$x^{2}$- 200 = 25

$x^{2}$ = 25 + 200

x = - 15

x = 15

Ответ: ±15.

б) $frac{3x+4}{x^{2}-16} = frac{x^{2}}{x^{2}-16}$

$frac{3x+4}{x^{2}-16} = frac{x^{2}}{x^{2}-16}$, x ≠ 4, x ≠ -4

3x + 4 - $x^{2}$ = 0

$-x^{2}$ - 3x -4 = 0

x = $frac{-(-3)±sqrt{(-3)^{2}-4*1*(-4) } }{2*1}$

x = $frac{3±sqrt{9+16} }{2}$

x =$frac{3-+sqrt{25}} {2}$

x =$frac{3+5}{2}$

x = $frac{3-5}{2}$

x = 4, x = -1, x ≠ 4, x ≠ -4

Ответ: x = -1

в) $frac{3}{x-5} +frac{8}{x}$ = 2

$frac{3}{x-5} +frac{8}{x}$ = 2, x ≠ 5, x ≠ 0

$frac{3}{x-5}+frac{8}{x}$ - 2 = 0

$frac{3x+8x-40-2x^{2}+10x}{x(x-5)}$ = 0

$frac{21x-40-2x^{2}}{x(x-5)}$  = 0

21x - 40 - 2x^{2} = 0

$-2x^{2}$ + 21x - 40 = 0

$2x^{2}$ - 21x + 40 = 0

x = $frac{21 ±sqrt{441-320}}{4}$

x = $frac{21±11}{4}$

x = $frac{21+11}{4}$

x = $frac{21-11}{4}$

x =8, x = 2,5, x ≠ 5, x ≠ 0.

Ответ: x = 2,5, x = 8